What is the proper set up to find how much heat in kJ is released when 147 grams of NO2 (g) is dissolved in excess water? 3 NO2 (g) + H2O (g) ---> 2 HNO3 (aq) + NO (g) AHrxn = -138 kJ 1 mol NO2 -138 KJ 147 g NO2 x 46.01 g NO2 * 1 mol NO2

Chemistry: Principles and Reactions
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Chapter8: Thermochemistry
Section: Chapter Questions
Problem 11QAP: When one mol of KOH is neutralized by sulfuric acid, q=56 kJ. (This is called the heat of...
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What is the proper set up to find how much heat in kJ is
released when 147 grams of NO2 (g) is dissolved in
excess water?
3 NO2 (g) + H20 (g) ---> 2 HNO3 (aq) + NO (g)
AHrxn = -138 kJ
1 mol NO2
-138 KJ
147 g NO2 x
46.01 g NO2“ 1 mol NO2
1 mol NO, -138 KJ
147 g NO, x
22.4 g NO2"1 mol NO2
1 mol NO2
-138 KJ
147 g NO, x
22.4 g NO2 * 3 mol NO2
1 mol NO2
-138 KJ
147 g NO2 x
46.01 g NO, * 3 mol NO2
Transcribed Image Text:What is the proper set up to find how much heat in kJ is released when 147 grams of NO2 (g) is dissolved in excess water? 3 NO2 (g) + H20 (g) ---> 2 HNO3 (aq) + NO (g) AHrxn = -138 kJ 1 mol NO2 -138 KJ 147 g NO2 x 46.01 g NO2“ 1 mol NO2 1 mol NO, -138 KJ 147 g NO, x 22.4 g NO2"1 mol NO2 1 mol NO2 -138 KJ 147 g NO, x 22.4 g NO2 * 3 mol NO2 1 mol NO2 -138 KJ 147 g NO2 x 46.01 g NO, * 3 mol NO2
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